Basic Trigonometric Function (amplitude, period)

I've got to find the period, amplitude and graph a trig function similar to the one below and I'm not sure if I'm on the right track. I haven't posted the exact question so if something doesn't look right with the function that's why.

From that I think it's:

Amplitude: 3

Period: 1 ie

Is raised up 8 on the y/sin axis so the max and minimum values are 11 and 5 with nothing to shift it right so graphing will pass through the point (0, 11)

Does that all seem right?

And what changes by there being a minus sign between the 8 and 3 rather than an addition?
Would the y axis start at the minimum point (5) rather than the maximum (11)?

Re: Basic Trigonometric Function (amplitude, period)

Originally Posted by Welkin

I've got to find the period, amplitude and graph a trig function similar to the one below and I'm not sure if I'm on the right track. I haven't posted the exact question so if something doesn't look right with the function that's why.

From that I think it's:

Amplitude: 3

Period: 1 ie

Is raised up 8 on the y/sin axis so the max and minimum values are 11 and 5 with nothing to shift it right so graphing will pass through the point (0, 11)

Does that all seem right?

And what changes by there being a minus sign between the 8 and 3 rather than an addition?
Would the y axis start at the minimum point (5) rather than the maximum (11)?

Re: Basic Trigonometric Function (amplitude, period)

You had me worried I was lost again.... well more lost

And it does start (or pass through) at (0, 5) doesn't it?
and I should read it as -3cos?

* Actually I'm still lost with this one, I thought that -cos means the y(sin) intercept is the low point rather than the high point but rereading my question here it also states that t = 0 corresponds with the temp being maxed (maxima?) which would be (0, 11) for the above example and so I'm still lost as to how the negative cos affects the graph?

Thanks for the help I'll add another one here doing the reverse:

If cycles through twice in an hour, has a maximum of 100 and a minimum of 50. = 0 corresponds to the maximum.

Then with the x (cos) axis representing minutes, I think the function would be:

Re: Basic Trigonometric Function (amplitude, period)

Originally Posted by Welkin

You had me worried I was lost again.... well more lost

And it does start (or pass through) at (0, 5) doesn't it?
and I should read it as -3cos?

* Actually I'm still lost with this one, I thought that -cos means the y(sin) intercept is the low point rather than the high point but rereading my question here it also states that t = 0 corresponds with the temp being maxed (maxima?) which would be (0, 11) for the above example and so I'm still lost as to how the negative cos affects the graph?

The f-intercept (read as y-intercept) is indeed (0, 5). The -3 is no problem...it simply means that we use -cos() instead of cos(). So the function moves from f = 5 to f = 11. Thus the amplitude is still 3.